The Practice of Statistics Daniel S. Yates The Practice of Statistics Third Edition Chapter 9: 9.3 Sample Means Copyright © 2008 by W. H. Freeman & Company

Essential Questions Given the means and standard deviation of a population, how do you calculate the means and standard deviation for the sampling distribution of a sample mean? What is the shape of the sampling distribution of a sample mean drawn from a population that has a Normal distribution? What is the central limit theorem? How do you use the central limit theorem to solve probability problems for the sampling distribution of a sample mean?

Sampling Distribution of The sampling distribution of is the distribution of the values of in all possible samples of the same size from the population. Two key facts: Means of random samples are less variable than individual observations Means of random samples are more Normal than individual observations.

The behavior of in repeated samples: The sample mean is an unbiased estimator of population mean The values of is less spread out for larger samples. The variability decreases at the rate of . The standard deviation of the sampling distribution of , can only be used when the population is at least 10 times as large as the sample. These facts about the means and standard deviation of are true regardless what the population distribution looks like.

The Shape of the Sampling Distribution of

Example – Young Women’s Heights The height of young women varies approximately according to the N(64.5, 2.5) distribution. If we choose one young woman at random, the heights we get in repeated choices follows this distribution. This distribution of the population is also the distribution of one observation chosen at random. We think of the population distribution as a distribution of probabilities.

Example – Young Women’s Heights The height of young women varies approximately according to the N(64.5, 2.5) distribution. What is the probability that a randomly selected young woman is taller than 66.5 inches? Using Table A, P(X > 66.5) = 1 – P(Z < 0.80) = 1 – 0.7661 = 0.2119 Using a calculator, P(X > 66.5) = NORMCDF(0.80, 9999) = 0.2119

Example – Young Women’s Heights The height of young women varies approximately according to the N(64.5, 2.5) distribution. Now, let’s take a SRS of 10 young women from this population and compute the means for this sample. Find the probability that the means of the sample will be greater than 66.5.

Fact: the average of several observations are less variable than individual observation. This fact leads to the central limit theorem.

The Central Limit Theorem and Non-normal Populations As sample size, n, increases the distribution of sample means becomes more normal.

More about the Central Limit Theorem. The Central Limit Theorem can safely be applied when n exceeds 30. If n is large or the population distribution is normal, the standardized variable has (approximately) a standard normal (z) distribution.

Example Servicing Air-conditioners The time that a technician requires to perform preventive maintenance on a air-conditioner is governed by an exponential density curve. The mean time is 1 hour and the standard deviation is 1 hour. Your company has a contract to maintain 70 units in an apartment building. You must schedule technicians’ time for a visit to this building. Is it safe to budget an average of 1.1 hours for each unit? Or should you budget an average of 1.25 hours? To determine whether it is safe to budget 1.1 hours, on the average, the probability we want is

Example Servicing Air-conditioners The time that a technician requires to perform preventive maintenance on a air-conditioner is governed by an exponential density curve. The mean time is 1 hour and the standard deviation is 1 hour. Your company has a contract to maintain 70 units in an apartment building. You must schedule technicians’ time for a visit to this building. Is it safe to budget an average of 1.1 hours for each unit? Or should you budget an average of 1.25 hours? To determine whether it is safe to budget 1.25 hours, on the average, the probability we want is Your turn – work this part of the problem. Based on your analysis, what is your decision?

Example A food company sells “18 ounce” boxes of cereal. Let x denote the actual amount of cereal in a box of cereal. Suppose that x is normally distributed with µ = 18.03 ounces and  = 0.05. a) What is the probability that a box will contain less than 18 ounces?

Example - continued b) A case consists of 24 boxes of cereal. What is the probability that the mean amount of cereal (per box in a case) is less than 18 ounces? The central limit theorem states that the distribution of is normally distributed so